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111. 二叉树的最小深度

English Version

题目描述

给定一个二叉树，找出其最小深度。

最小深度是从根节点到最近叶子节点的最短路径上的节点数量。

说明：叶子节点是指没有子节点的节点。

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：2

示例 2：

输入：root = [2,null,3,null,4,null,5,null,6]
输出：5

提示：

树中节点数的范围在 [0, 10⁵] 内
-1000 <= Node.val <= 1000

解法

方法一：递归

递归的终止条件是当前节点为空，此时返回 $0$；如果当前节点左右子树有一个为空，返回不为空的子树的最小深度加 $1$；如果当前节点左右子树都不为空，返回左右子树最小深度的较小值加 $1$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        if root.left is None:
            return 1 + self.minDepth(root.right)
        if root.right is None:
            return 1 + self.minDepth(root.left)
        return 1 + min(self.minDepth(root.left), self.minDepth(root.right))

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null) {
            return 1 + minDepth(root.right);
        }
        if (root.right == null) {
            return 1 + minDepth(root.left);
        }
        return 1 + Math.min(minDepth(root.left), minDepth(root.right));
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) {
            return 0;
        }
        if (!root->left) {
            return 1 + minDepth(root->right);
        }
        if (!root->right) {
            return 1 + minDepth(root->left);
        }
        return 1 + min(minDepth(root->left), minDepth(root->right));
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDepth(root *TreeNode) int {
	if root == nil {
		return 0
	}
	if root.Left == nil {
		return 1 + minDepth(root.Right)
	}
	if root.Right == nil {
		return 1 + minDepth(root.Left)
	}
	return 1 + min(minDepth(root.Left), minDepth(root.Right))
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDepth(root: TreeNode | null): number {
    if (root == null) {
        return 0;
    }
    const { left, right } = root;
    if (left == null) {
        return 1 + minDepth(right);
    }
    if (right == null) {
        return 1 + minDepth(left);
    }
    return 1 + Math.min(minDepth(left), minDepth(right));
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        if node.left.is_none() {
            return 1 + Self::dfs(&node.right);
        }
        if node.right.is_none() {
            return 1 + Self::dfs(&node.left);
        }
        1 + Self::dfs(&node.left).min(Self::dfs(&node.right))
    }

    pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        Self::dfs(&root)
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function (root) {
    if (!root) {
        return 0;
    }
    if (!root.left) {
        return 1 + minDepth(root.right);
    }
    if (!root.right) {
        return 1 + minDepth(root.left);
    }
    return 1 + Math.min(minDepth(root.left), minDepth(root.right));
};

C

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

#define min(a, b) (((a) < (b)) ? (a) : (b))

int minDepth(struct TreeNode* root) {
    if (!root) {
        return 0;
    }
    if (!root->left) {
        return 1 + minDepth(root->right);
    }
    if (!root->right) {
        return 1 + minDepth(root->left);
    }
    int left = minDepth(root->left);
    int right = minDepth(root->right);
    return 1 + min(left, right);
}

方法二：BFS

使用队列实现广度优先搜索，初始时将根节点加入队列。每次从队列中取出一个节点，如果该节点是叶子节点，则直接返回当前深度；如果该节点不是叶子节点，则将该节点的所有非空子节点加入队列。继续搜索下一层节点，直到找到叶子节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        q = deque([root])
        ans = 0
        while 1:
            ans += 1
            for _ in range(len(q)):
                node = q.popleft()
                if node.left is None and node.right is None:
                    return ans
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        int ans = 0;
        while (true) {
            ++ans;
            for (int n = q.size(); n > 0; n--) {
                TreeNode node = q.poll();
                if (node.left == null && node.right == null) {
                    return ans;
                }
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) {
            return 0;
        }
        queue<TreeNode*> q{{root}};
        int ans = 0;
        while (1) {
            ++ans;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                if (!node->left && !node->right) {
                    return ans;
                }
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDepth(root *TreeNode) (ans int) {
	if root == nil {
		return 0
	}
	q := []*TreeNode{root}
	for {
		ans++
		for n := len(q); n > 0; n-- {
			node := q[0]
			q = q[1:]
			if node.Left == nil && node.Right == nil {
				return
			}
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDepth(root: TreeNode | null): number {
    if (!root) {
        return 0;
    }
    const q = [root];
    let ans = 0;
    while (1) {
        ++ans;
        for (let n = q.length; n; --n) {
            const node = q.shift();
            if (!node.left && !node.right) {
                return ans;
            }
            if (node.left) {
                q.push(node.left);
            }
            if (node.right) {
                q.push(node.right);
            }
        }
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function (root) {
    if (!root) {
        return 0;
    }
    const q = [root];
    let ans = 0;
    while (1) {
        ++ans;
        for (let n = q.length; n; --n) {
            const node = q.shift();
            if (!node.left && !node.right) {
                return ans;
            }
            if (node.left) {
                q.push(node.left);
            }
            if (node.right) {
                q.push(node.right);
            }
        }
    }
};

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README.md

README.md

111. 二叉树的最小深度

题目描述

解法

方法一：递归

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

C

方法二：BFS

Python3

Java

C++

Go

TypeScript

JavaScript

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111. 二叉树的最小深度

题目描述

解法

方法一：递归

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

C

方法二：BFS

Python3

Java

C++

Go

TypeScript

JavaScript